Vectors in three dimensions - TCE Mathematics Specialised (Tasmania)

Syllabus dot point

“Use 3D vectors with dot and cross products to find angles, projections, lines and planes.”

TCESpecialist MathematicsUnit 39 min readUpdated 2 Jun 2026

Quick answer

Work with vectors in 3D as $\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}$ . Use the dot product for angles and projections, the cross product for a perpendicular vector and areas, and combine these to write vector equations of lines and planes.

What this dot point is asking

In three dimensions a vector has three components, $\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}$ , often written as a column $(a_1, a_2, a_3)$ . The basic operations carry over from two dimensions, but two new ideas become essential: the cross product, and the description of lines and planes.

Magnitude and unit vectors

The magnitude (length) is

|\mathbf{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}.

A unit vector in the direction of $\mathbf{a}$ is $\hat{\mathbf{a}} = \dfrac{\mathbf{a}}{|\mathbf{a}|}$ . Unit vectors are useful whenever you only care about direction, for example when writing the direction of a line.

The dot (scalar) product

\mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 = |\mathbf{a}|\,|\mathbf{b}|\cos\theta,

where $\theta$ is the angle between the vectors. Rearranging gives the angle:

\cos\theta = \frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}|\,|\mathbf{b}|}.

Two non-zero vectors are perpendicular exactly when $\mathbf{a}\cdot\mathbf{b} = 0$ . The scalar projection of $\mathbf{a}$ onto $\mathbf{b}$ is $\dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|}$ , and the vector projection is $\dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|^2}\,\mathbf{b}$ .

The cross (vector) product

The cross product produces a vector perpendicular to both inputs:

\mathbf{a}\times\mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} = (a_2 b_3 - a_3 b_2)\mathbf{i} - (a_1 b_3 - a_3 b_1)\mathbf{j} + (a_1 b_2 - a_2 b_1)\mathbf{k}.

Its magnitude is $|\mathbf{a}\times\mathbf{b}| = |\mathbf{a}|\,|\mathbf{b}|\sin\theta$ , which equals the area of the parallelogram spanned by $\mathbf{a}$ and $\mathbf{b}$ . Note that $\mathbf{a}\times\mathbf{b} = -(\mathbf{b}\times\mathbf{a})$ , so order matters.

Dot versus cross

The dot product returns a scalar and is zero when vectors are perpendicular. The cross product returns a vector and is the zero vector when vectors are parallel. Pick the one that matches what the question wants: an angle or projection (dot), or a normal direction or area (cross).

Lines and planes

The vector equation of a line through point $\mathbf{r}_0$ with direction $\mathbf{d}$ is

\mathbf{r} = \mathbf{r}_0 + t\mathbf{d}, \qquad t \in \mathbb{R}.

The equation of a plane with normal $\mathbf{n} = (a,b,c)$ through point $\mathbf{r}_0$ is $\mathbf{n}\cdot(\mathbf{r} - \mathbf{r}_0) = 0$ , which expands to the Cartesian form $ax + by + cz = d$ where $d = \mathbf{n}\cdot\mathbf{r}_0$ . To find a normal to a plane through three points, take the cross product of two vectors lying in the plane.

A plane through three points

Find the Cartesian equation of the plane through $A(1,0,2)$ , $B(3,1,1)$ and $C(2,2,0)$ .

Step 1: Form two vectors lying in the plane

A plane is determined by a point and a normal. To get the normal we need two independent vectors that lie in the plane. We compute these by subtracting position vectors of the given points:

\overrightarrow{AB} = B - A = (3-1,\; 1-0,\; 1-2) = (2,1,-1), \qquad \overrightarrow{AC} = C - A = (1,2,-2).

Step 2: Compute the normal via the cross product

The cross product $\overrightarrow{AB} \times \overrightarrow{AC}$ gives a vector perpendicular to both, hence perpendicular to the plane:

\mathbf{n} = \overrightarrow{AB}\times\overrightarrow{AC} = \big( (1)(-2)-(-1)(2),\; -[(2)(-2)-(-1)(1)],\; (2)(2)-(1)(1) \big).

Component by component: the $\mathbf{i}$ term is $-2 + 2 = 0$ ; the $\mathbf{j}$ term is $-(-4 + 1) = 3$ ; the $\mathbf{k}$ term is $4 - 1 = 3$ . So $\mathbf{n} = (0, 3, 3)$ , which we simplify to $(0,1,1)$ .

Step 3: Write the Cartesian equation using one known point

Substituting point $A(1,0,2)$ into $\mathbf{n}\cdot\mathbf{r} = d$ gives the constant:

d = \mathbf{n}\cdot\mathbf{r}_0 = (0)(1) + (1)(0) + (1)(2) = 2.

The plane is therefore:

y + z = 2.

Check with $B$ : $1 + 1 = 2$ . Check with $C$ : $2 + 0 = 2$ . Both satisfy it, confirming the equation is correct.

Final answer: $y + z = 2$ .

Forgetting the cross product is anticommutative

Students often treat $\mathbf{a}\times\mathbf{b}$ and $\mathbf{b}\times\mathbf{a}$ as equal. They are not: $\mathbf{b}\times\mathbf{a} = -(\mathbf{a}\times\mathbf{b})$ , so the normal points the opposite way. For a plane equation either direction works, but in problems about orientation, torque or a right-hand rule it changes the answer. Always keep the order the question implies and state which vectors you crossed.

Putting it together

Many exam questions chain these tools: find a direction with subtraction, a normal with a cross product, an angle with a dot product, then assemble a line or plane equation. Lay your working out in clear named steps so each result is easy to reuse later in the question.

Exam-style questions

Questions in the style of TASC exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice6 marks

The point

P(1, -2, 3)

lies on the plane

\Pi: 3x + 4y - 2z = d

. (a) Determine

d

. (b) Write the line

L

through

P

parallel to direction

(2, \mu, -1)

in parametric form. (c) If

L

lies in

\Pi

, determine

\mu

Show worked answer →

(a) Substitute $P$ into the plane equation: $3(1) + 4(-2) - 2(3) = 3 - 8 - 6 = -11$ . So $d = -11$ . (2 marks)

(b) A line through $P$ with direction $(2, \mu, -1)$ is $(x, y, z) = (1, -2, 3) + t(2, \mu, -1)$ , that is $x = 1 + 2t$ , $y = -2 + \mu t$ , $z = 3 - t$ . (2 marks)

(c) If $L$ lies in the plane, its direction must be perpendicular to the plane's normal $(3, 4, -2)$ . Set the dot product to zero: $3(2) + 4\mu + (-2)(-1) = 6 + 4\mu + 2 = 0$ , so $4\mu = -8$ and $\mu = -2$ . Because $P$ already lies on the plane, this condition is sufficient. (2 marks)

Practice5 marks

Planes

\Pi_1: x + 2y - 3z = 4

and

\Pi_2: 2x + 4y + cz = d

. (a) Show

P(3, 2, 1)

lies on

\Pi_1

and

Q\left(c, \dfrac{d}{4}, -2\right)

lies on

\Pi_2

. (b) For what value of

c

are the planes parallel? (c) With this

c

and

d = 8

, prove

Q

also lies on

\Pi_1

Show worked answer →

(a) For $P$ on $\Pi_1$ : $3 + 2(2) - 3(1) = 3 + 4 - 3 = 4$ , true. For $Q$ on $\Pi_2$ : $2c + 4\left(\dfrac{d}{4}\right) + c(-2) = 2c + d - 2c = d$ , true. (2 marks)

(b) Two planes are parallel when their normals are scalar multiples. The normal of $\Pi_1$ is $(1, 2, -3)$ and the normal of $\Pi_2$ is $(2, 4, c)$ . Matching $(2, 4, c) = 2(1, 2, -3)$ gives $c = -6$ . (1 mark)

(c) With $c = -6$ and $d = 8$ , $Q = (-6, 2, -2)$ . Substitute into $\Pi_1$ : $(-6) + 2(2) - 3(-2) = -6 + 4 + 6 = 4$ , which equals the right side, so $Q$ lies on $\Pi_1$ . (2 marks)

Practice6 marks

Q

is the plane

4x - 3y + 6z = 36

. (a) Show

A(3, 2, 5)

lies on

Q

. (b) Determine the equation of the line joining

A

B(1, 2, -3)

. (c) Find the equation of the plane parallel to

Q

that contains

B

Show worked answer →

(a) Substitute $A$ : $4(3) - 3(2) + 6(5) = 12 - 6 + 30 = 36$ , which matches, so $A$ lies on $Q$ . (1 mark)

(b) The direction $\overrightarrow{AB} = B - A = (1 - 3, 2 - 2, -3 - 5) = (-2, 0, -8)$ , which simplifies to $(1, 0, 4)$ . The line is $(x, y, z) = (3, 2, 5) + t(1, 0, 4)$ , that is $x = 3 + t$ , $y = 2$ , $z = 5 + 4t$ . (3 marks)

(c) A parallel plane has the same normal $(4, -3, 6)$ , so it has the form $4x - 3y + 6z = k$ . Substitute $B(1, 2, -3)$ : $4 - 6 - 18 = -20$ . The plane is $4x - 3y + 6z = -20$ . (2 marks)

Practise this

Sources & how we know this

Mathematics Specialised Level 4 (MTS415118) course document (2018 onwards) — TASC